Question

我需要输出一个JSON对象供iphone使用

我能输出像

{"feed":{"id":"1","player":"player1"}}
{"feed":{"id":"1","player":"player2"}}
{"feed":{"id":"2","player":"player3"}}

代码：

$query = "SELECT id,player FROM MyVideos";
$result = mysql_query($query,$link) or die('Errant query: '.$query);

$players[] = array();
if(mysql_num_rows($result)){
while($player = mysql_fetch_assoc($result)){
$players[] = array('player'=>$player);
echo json_encode(array("feed"=>$player));            
}
}

但是我需要输出像这样的东西

{"feed":
{"id":"1","player":"player1"},
{"id":"1","player":"player2"},
{"id":"2","player":"player3"}
}

任何人都可以帮助我。

谢谢，

Answer 1

您发布的输出无效JSON，您需要在feed中的项目周围添加括号：

{"feed": [
    {"id":"1","player":"player1"},
    {"id":"1","player":"player2"},
    {"id":"2","player":"player3"}
]}

您应该循环搜索结果并构建一个Feed项目数组，然后立即输出所有内容，如下所示：

$feed_items = array();

if (mysql_num_rows($result)) {
    while ($player = mysql_fetch_assoc($result)){
        $feed_items[] = $player;
    }
}

echo json_encode(array("feed" => $feed_items));

使用PHP的JSON输出

1 个答案: